Question

How many moles of O₂ would be required to generate 13.0 mol of NO₂ in the reaction below assuming the reaction has only 80.0% yield?

Answer 1

Answer:

Chemical reaction:

• 2 NO (g) + O₂ (g) → 2 NO₂ (g)

2 moles of NO₂ is produced from O₂ moles = 1

Therefore, 2 moles of NO₂ is produced from O₂ moles = $\dfrac{1}{2} \times 13 = \bf{6.5 mol}$

For 80 % yield no. of moles of O₂ = 6.5 mol

Then for 100% yield no. of moles of O₂ =$\dfrac{6.5}{80\%} =\dfrac{6.5}{\frac{80}{100}} =\dfrac{6.5}{80} \times 100 = \dfrac{650}{80} = \bf 8.125\;mol$

Answer 2

Answer:

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